Find The Last 2 Digits

any number 10! or greater we can ignore; it's guaranteed to end in at least two zeroes, since within the factorial we are multiplying 2 x 5 x 10 to get 100.

this just leaves us with (0! + 5!) mod 100, which we can easily calculate and simplify to 121 mod 100 = 21.

First look at numbers greater than or equal to 10! We have two fives and atleast two two's. This means that any number above 9! can be written as $2\times 2\times 5\times 5\times A$ where A is the rest of the multiplied integers. Thus the number will always have two trailing zeroes.

Now we only need to consider $0! +5!$

This is very simply $120+1( mod 100)$ Thus the answer is $\boxed {21}$

After 9! Every no. Has last two digits zero.so we are left with 0! And 5! . Which means 0! + 5! Will give last two digits of the complete addition. 0! + 5! = 1+ 120 =121 So the last 2 digits are 21

15! to 100! have zero on the last 2 digits.0!=1 + 5!=120==121.ans=21

0!=1 ; 5!=120. rest all the numbers have 0 in their ones and tense place. Therefore, last 2 digits will be 21. because 0! + 5! = 121. hence last 2 digits are 21

we know that after 5! all no's have 00 as their2 last digits :. 0!=1&5!=120 120+1=121 but we are asked last two digits so they are 21

I guess 21 and it was correct

0!=1

5!=120

10! and other terms end up with 0(zero) at unit and at tenth place.

So 0! + 5!= 121

Hence the answer is 21

0! + 5! + 10! ...= 0! +5! +10!(k)=21 +10!(k) 10! has unit and tens digits as zero. So, soln is 21

We have have that $(5j)!$ for $j$ at least 2 is congruent to 0 modulo 100. So, it suffices to find $0! + 5! \pmod {100}$ which is easily seen to be 21.

solution: last two digits of 0!=01 5!=20 10!=00 . . 100!=00

so the answer is 21

because every term has 100 expect first two terms and first term =1 and second one =20 then adding last two digit 20+ 1=21